Phys101- lec9-recording on 06 March 2025

Introduction to Moment of Inertia

• The moment of inertia is crucial for understanding the rotational dynamics of objects.

• A rotation axis not going through the center of mass requires special consideration to determine the moment of inertia.

Parallel Axis Theorem

• The Parallel Axis Theorem is used to calculate the moment of inertia of an object when the axis of rotation is not through its center of mass.

• Statement of the Theorem: [ I = I_{cm} + m imes d^2 ]

  • Where:

    • ( I ) = moment of inertia about any parallel axis

    • ( I_{cm} ) = moment of inertia about the center of mass

    • ( m ) = total mass of the object

    • ( d ) = distance from the center of mass to the new axis

• Example with a uniform rod:

  • If the rod's center of mass is at ( l/2 ), then ( d = l/2 ) for rotation about one end.

  • Moment of inertia through the center of mass: ( I_{cm} = \frac{1}{12} m l^2 )

  • Updated calculation yields ( I = \frac{1}{3} m l^2 ).

Torque and Its Calculation

• Torque (( \tau )) is the rotational equivalent of force. It is defined as:

  • ( \tau = r \times f )

• The direction of torque is determined using the right-hand rule:

  • Align fingers along the radial vector (r), curl towards the force direction (f), thumb points in the torque direction.

• Vector Notation:

  • The order matters: ( \tau
    eq f \times r )

Torque Calculation Example

• To compute torque from two vectors:

  • Use a 3x3 determinant format involving unit vectors ( i, j, k ).

  • Example determinant: [ \begin{vmatrix} i & j & k \ r_x & r_y & r_z \ f_x & f_y & f_z \end{vmatrix} ]

• Important to calculate each component of the torque vector correctly and to maintain sign conventions.

Application of Newton’s Laws in Rotational Motion

• Equations of motion for linear and rotational dynamics link through the tensions in string systems:

  • For a mass sliding on a surface, apply Newton's second law:

    • ( m_1 g - T_1 = m_1 a ) for the mass in free fall.

    • ( T_2 = m_2 a ) for the sliding mass.

• The sum of torques relates to angular acceleration:

  • ( \tau = I \alpha )

  • Relate angular acceleration to linear acceleration using ( a = \alpha r ).

Example Problem: Moment of Inertia of a Pulley

• Given:

  • Mass ( m_1 ) hanging vertically, mass ( m_2 ) on a surface, and a pulley of radius ( r ).

  • Free body diagram includes forces: gravitational force and tension.

• Establish free body equations for each mass:

  • Combine forces to derive equations for tensions in strings.

• Solve for the moment of inertia using:

  • ( I = \frac{r^2 (T_1 - T_2)}{a} ) where ( T_1 ) and ( T_2 ) are tensions derived from their respective linear equations.

• Example results yielded an inertia of ( 0.2 , ext{kg m}^2 ).

Conclusion

• Understanding the moment of inertia, parallel axis theorem, and torque calculations is fundamental in addressing problems involving rotation and dynamics.